Relations of $S^2 V$ and heighest weight representations of Lie algebras.

Question

Let $V$ be the natural representation of $sl_n$. Then $V = V(\omega_1)$, where $\omega_1$ is the first fundamental weight. We have $\Lambda^2 V = V(\omega_2)$. Is $S^2 V = V(\lambda)$ for some weight $\lambda$? Thank you very much.

Answer 1

$S^2(V)$ contains a highest weight vector of weight $2\omega_1$ and this weight space is 1 dimensional. Also $dim(S^2(V))=\frac{n(n+1)}{2}$

$dim(V(2\omega_1))=\prod_{j=2}^{n}\frac{a_1+j-1}{j-1}=\frac{n(n+1)}{2}$. Since $S^2(V)$ contains a highest weight vector of weight $2\omega_1$, $V(2\omega_1)$ is a subrepresentation of $S^2(V)$. As a result, $S^2(V)\cong V(2\omega_1)$.